What is the extreme f(x)=(sqrt(2))/2 x+sin(x) ?

The extreme f(x)=(sqrt(2))/2 x+sin(x) is

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extreme f(x)=(sqrt(2))/2 x+sin(x)

extreme points

f (x) = \frac{2}{2} x + sin (x)

Maximum (\frac{3 π}{4} + 2 πn, \frac{3 π + 8 πn}{4 2} + sin (\frac{3 π}{4} + 2 πn)), Minimum (\frac{5 π}{4} + 2 πn, \frac{5 π + 8 πn}{4 2} + sin (\frac{5 π}{4} + 2 πn))

Solution steps

Find the critical points:

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Domain of

\frac{2}{2} x + sin (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{3 π}{4} + 2 πn,

Decreasing

: \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn,

Increasing

: \frac{5 π}{4} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{3 π}{4} + 2 πn

into

\frac{2}{2} x + sin (x) \Rightarrow y = \frac{3 π + 8 πn}{4 2} + sin (\frac{3 π}{4} + 2 πn)

Maximum (\frac{3 π}{4} + 2 πn, \frac{3 π + 8 πn}{4 2} + sin (\frac{3 π}{4} + 2 πn))

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

\frac{2}{2} x + sin (x) \Rightarrow y = \frac{5 π + 8 πn}{4 2} + sin (\frac{5 π}{4} + 2 πn)

Minimum (\frac{5 π}{4} + 2 πn, \frac{5 π + 8 πn}{4 2} + sin (\frac{5 π}{4} + 2 πn))

Maximum (\frac{3 π}{4} + 2 πn, \frac{3 π + 8 πn}{4 2} + sin (\frac{3 π}{4} + 2 πn)), Minimum (\frac{5 π}{4} + 2 πn, \frac{5 π + 8 πn}{4 2} + sin (\frac{5 π}{4} + 2 πn))

f (x) = x^{3} + 3 x^{2} - 9 x + 8

f (x) = x^{2} lo g_{5} (x)

f (x) = (x + 2)^{3} (x - 3)

f (x) = (x - 1) (x - 6)^{3} + 6, 1 \leq x \leq 4

What is the extreme f(x)=(sqrt(2))/2 x+sin(x) ?
The extreme f(x)=(sqrt(2))/2 x+sin(x) is